1. **Problem Statement:** We want to find the smallest eigenvalue and its corresponding eigenvector of the matrix $$A = \begin{bmatrix} 5 & 2 & -1 \\ 2 & 1 & 1 \\ -3 & 3 & 4 \end{bmatrix}$$ using the Inverse Power Method with initial vector $$v_0 = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$$ and tolerance $$\varepsilon = 0.01$$.
2. **Inverse Power Method Overview:** The Inverse Power Method finds the eigenvalue closest to zero (smallest in magnitude) by iteratively applying the inverse of the matrix to a vector.
3. **Key Formula:**
$$
v_{k+1} = \frac{A^{-1} v_k}{\|A^{-1} v_k\|}
$$
The Rayleigh quotient approximation for the eigenvalue is:
$$
\lambda_k = \frac{v_k^T A v_k}{v_k^T v_k}
$$
4. **Steps:**
- Start with initial vector $$v_0$$.
- Solve $$A y = v_k$$ for $$y$$.
- Normalize $$y$$ to get $$v_{k+1}$$.
- Compute $$\lambda_k$$ using the Rayleigh quotient.
- Repeat until $$|\lambda_{k+1} - \lambda_k| < \varepsilon$$.
5. **Iteration 1:**
Solve $$A y = v_0$$:
$$
\begin{bmatrix} 5 & 2 & -1 \\ 2 & 1 & 1 \\ -3 & 3 & 4 \end{bmatrix} y = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}
$$
Using matrix inversion or linear solver, we find:
$$
y \approx \begin{bmatrix} 0.176 \\ 0.294 \\ 0.118 \end{bmatrix}
$$
Normalize $$y$$:
$$
\|y\| = \sqrt{0.176^2 + 0.294^2 + 0.118^2} \approx 0.36
$$
$$
v_1 = \frac{y}{0.36} \approx \begin{bmatrix} 0.49 \\ 0.82 \\ 0.33 \end{bmatrix}
$$
Compute $$\lambda_1$$:
$$
\lambda_1 = \frac{v_1^T A v_1}{v_1^T v_1} \approx \frac{\begin{bmatrix}0.49 & 0.82 & 0.33\end{bmatrix} \begin{bmatrix}5 & 2 & -1 \\ 2 & 1 & 1 \\ -3 & 3 & 4 \end{bmatrix} \begin{bmatrix}0.49 \\ 0.82 \\ 0.33 \end{bmatrix}}{1} \approx 0.91
$$
6. **Iteration 2:**
Solve $$A y = v_1$$:
$$
y \approx \begin{bmatrix} 0.44 \\ 0.68 \\ 0.27 \end{bmatrix}
$$
Normalize $$y$$:
$$
\|y\| \approx 0.82
$$
$$
v_2 = \frac{y}{0.82} \approx \begin{bmatrix} 0.54 \\ 0.83 \\ 0.33 \end{bmatrix}
$$
Compute $$\lambda_2$$:
$$
\lambda_2 \approx 0.91
$$
7. **Convergence Check:**
$$|\lambda_2 - \lambda_1| = |0.91 - 0.91| = 0 < 0.01$$, so we stop.
8. **Final Answer:**
The smallest eigenvalue is approximately $$\boxed{0.91}$$ with corresponding eigenvector $$\boxed{\begin{bmatrix} 0.54 \\ 0.83 \\ 0.33 \end{bmatrix}}$$.
This method efficiently finds the smallest eigenvalue by repeatedly solving linear systems and normalizing the vectors.
Inverse Power Method D606Ee
Step-by-step solutions with LaTeX - clean, fast, and student-friendly.